If {}^nC_{r-1}=36,{}^nC_r=84,and {}^nC_{r+1}= | vedclass

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(D) We are given the combinations: ${}^nC_{r-1}=36$,${}^nC_r=84$,and ${}^nC_{r+1}=126$.Taking the ratio $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{84}{36} =

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$81$

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(D) We are given the combinations: ${}^nC_{r-1}=36$,${}^nC_r=84$,and ${}^nC_{r+1}=126$.Taking the ratio $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}$.Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{7}{3}$ $\Rightarrow 3n-3r+3 = 7r$ $\Rightarrow 3n+3 = 10r$ (Equation $i$).Taking the ratio $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{126}{84} = \frac{3}{2}$.Using the formula $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}$,we get $\frac{n-r}{r+1} = \frac{3}{2}$ $\Rightarrow 2n-2r = 3r+3$ $\Rightarrow 2n-3 = 5r$ (Equation $ii$).From Equation $ii$,$r = \frac{2n-3}{5}$. Substituting this into Equation $i$:$3n+3 = 10 \left( \frac{2n-3}{5} ight)$ $\Rightarrow 3n+3 = 2(2n-3)$ $\Rightarrow 3n+3 = 4n-6$ $\Rightarrow n = 9$.Substituting $n=9$ into Equation $ii$: $2(9)-3 = 5r$ $\Rightarrow 15 = 5r$ $\Rightarrow r = 3$.Therefore,$nr^2 = 9 	imes (3)^2 = 9 	imes 9 = 81$.

If ${}^nC_{r-1}=36$,${}^nC_r=84$,and ${}^nC_{r+1}=126$,then the value of $nr^2$ is

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